# Why V rms instead of V average?

I'm looking at an equation for average power in a signal

$$p_{avg} = \frac{1}{R} v_{rms}^2$$

and wondering why it isn't

$$p_{avg} = \frac{1}{R} |v|_{avg}^2$$

• Because the square of the average is not always the average of the squares, not even for positive numbers. 0 and 10 average to 5, square that to get 25. But the average of their squares (0 and 100) is 50. Not even close! Why the square in the first place? Power is Voltage * current, but the current is itself proportional to the Voltage, so the power is proportional to the Voltage squared. Commented Sep 14, 2012 at 20:14
• The first equation isn't a mathematical definition or a physical law; it can be proven. Look how it is derived and your question will be answered. Commented Oct 7, 2020 at 13:10

Simple: the average of a sine is zero.

Power is proportional to voltage squared:

$$\ P = \dfrac{V^2}{R} \$$

so to get average power you calculate average voltage squared. That's what the RMS refers to: Root Mean Square: take the square root of the average (mean) of the squared voltage. You have to take the square root to get the dimension of a voltage again, since you first squared it.

This graph shows the difference between the two. The purple curve is the sine squared, the yellowish line the absolute value. The RMS value is $$\\sqrt{2}/2\$$ (i.e. $$\\frac{1}{\sqrt{2}}\$$), or about 0.71, the average value is $$\2/\pi\$$, or about 0.64, a difference of 10 %.

RMS gives you the equivalent DC voltage for the same power. If you would measure the resistor's temperature as a measure of dissipated energy you'll see that it's the same as for a DC voltage of 0.71 V, not 0.64 V.

edit
Measuring average voltage is cheaper than measuring RMS voltage however, and that's what cheaper DMMs do. They presume the signal is a sine wave, measure the rectified average and multiply the result by 1.11 (0.71/0.64) to get the RMS value. But the factor 1.11 is only valid for sinewaves. For other signals the ratio will be different. That ratio got a name: it's called the signal's form factor. For a 10 % duty cycle PWM signal the form factor will be $$\1/\sqrt{10}\$$, or about 0.316. That's a lot less than the sine's 1.11. DMMs which are not "True RMS" will give large errors for non-sinusoidal waveforms.

• To your first point, I edited my second equation to use the average absolute value, which is what I meant. What I'm not seeing is why the order of the two operations (average and square) matters. Average voltage squared, vs average squared voltage. Commented Sep 14, 2012 at 17:13
• Because of the square-law relationship the average of the power and the average of the voltage are two very different things. Commented Sep 14, 2012 at 17:27
• @RobN, the instantaneous power is $p(t) = v^2(t)/R$. The average power is the time average of $p(t)$. Thus, the average power is proportional to the time average of the squared voltage. Also, the order matters because the average of the squares is not equal to the square of the average. Commented Sep 14, 2012 at 19:33
• @RobN: The difference between average voltage squared $( \langle A \rangle + \langle B \rangle + \langle C \rangle)^{2}$ and average squared voltage ${\langle A^2 \rangle} + {\langle B^2 \rangle} + {\langle C^2 \rangle}$ is a matter in order of operations. In the first, you would have multiple cross terms which you will not have in the second. Commented Nov 14, 2018 at 1:04
• The prase "so to get average power you calculate average voltage squared" is wrong; that's the second equation in the question asked (which is the wrong one.) The equation of average power doesn't use the square of the average of the voltage, but instead the average of the square of voltage, as Kd_R's answer shows. Commented Oct 7, 2020 at 13:38

Now speaking in terms of equations:

$$\ P_{avg}= avg(P_{inst}) \$$

Now $$\P_{inst} = v(t) \cdot i(t)\$$ where $$\v(t)\$$ and $$\i(t)\$$ are instantaneous voltage and current resp. Hence

$$\ P_{inst} = \dfrac{(v(t))^2}{R} \$$

$$\ P_{avg} = avg \left(\dfrac{((v(t))^2}{R} \right) \$$

$$\ P_{avg} = \dfrac{(\sqrt{avg(v(t)^2)})^2}{R} \$$

$$\ P_{avg}= \dfrac{V_{rms}^2}{R} \$$

As RMS = $$\\sqrt{\text{average of squares of inst.}}\$$

• So? All you have presented is equations, without explanation or argument. This is not useful. Commented Nov 14, 2018 at 4:56
• Have only one question that still is not clear. Why everybody explains this via Power? Why not Current (I)? There is no square in Current, why same RMS applies for Current calculation? Commented May 14, 2022 at 21:18

The why is simple.

You want 1 W = 1 W.

Imagine a primitive heater, a 1 ohm resistor.

Consider 1 VDC into a 1 ohm resistor. Power consumption is obviously 1 W. Do that for one hour, and you burn one watt-hour, generating heat.

Now, instead of DC, you want to feed AC to the resistor, and produce the same heat. What AC voltage do you use?

It turns out that RMS voltage gives you the result you want.

THAT'S why RMS is defined the way it is, to make the power numbers come out right.

• This has hints of a useful answer, but it must be all but entirely rewritten to make them clear Commented Nov 14, 2018 at 4:57

Because the power equal to V^2/R so that you calculate the average of the squared voltages along the sinusoidal wave to get V^2avg. For simplicity we take the average of this mean then we can deal with it as we wish.

• This is essentially the key point, but it could be explained in a much better way. Commented Nov 14, 2018 at 4:57
• Average of the mean?? True, you use the square of the voltage to get equivalent power, BUT you take the square root of the mean squared voltage to get back to voltage. Commented Dec 18, 2022 at 19:46
• A typo :) it is the root of this mean Commented Jan 8 at 7:29

The answer is the reason given by John R. Strohm and the explanation is as follows: (requires a few additions to stevenvh's answer)

You see when you send a DC through a resistor and an AC wave through a resistor the resistor get heated up in both cases, but according to the equation for the average value the heating effect for ac should be 0 but its not why? This is because the when the electrons move in a conductor they strike atoms and this energy imparted to the atoms are consequently felt as heat, now AC does the same thing only the the electrons are moving in different directions but the energy transfer here is independent of the direction and so the conductor heats up all the same.

When we find the average value the ac components are cancelled out and hence fail to explain why the heat is generated but RMS equation rectifies that - as stevenvh says by taking the square and then the square root we are transposing the negative portion to the top of the axis such that the positive and negative portions don't cancel off.

This is why that we say that the average and the RMS values of a DC wave is the same.

The same applies to any real world signal (by this I mean imperfect - not pure AC) as Fourier series says that any wave can be replaced by a correct combination of sine and cosine waves and since the frequencies of the waves are higher (integer multiples of the base frequency) they too get cancelled out, isolating the DC component.

The above is the reason that we define RMS value as the equivalent value of DC that generates the same amount of heat as the AC wave.

Hope this helps.

PS: I know that the explanation for how heat is generated is quite ambiguous but I am at loss to find a better one, I went with it anyway because it helps convey the message

• There are some useful points here, but this is far too chatty; to be a good answer you must entirely re-write this in a factual way. Commented Nov 14, 2018 at 4:58

Where did the Vrms equation come from? Why does it work?

Our analysis need only consider a quarter of a cycle. The other three-quarters yield the same area under the curve, the same squares, the same power.

By definition we are looking for equivalent AC and DC power. Power is a function of the square of the voltage (Power=V²/R). Finding the square of a DC voltage is easy. How do we find the square of a voltage changing as a sine wave? You chop up the sine wave and square the voltage at each slice. Graphically, take each voltage slice and draw it as a square shooting out into the z-axis. Add up all the square areas then divide by the number of squares. That gives you the average area (mean of the squares) for that sine wave with a peak Vp.

Now that we have the average square area, we take the square root of it to get the voltage that gives us that average square area and... oh my god we've derived the Vrms formula!

The same question was bugging me. Looking on the web turned up answers which didn't ease the existential angst. The most common one was that the average of a full cycle of a sine wave is zero, so we use RMS instead. The implication of this line of thought seems to be that the square root of the average of the squares (RMS) of a series is equal to the average of the absolute values of the series, which is not the case.

$$\sqrt{\frac{1^2 + 2^2 + 3^2}{3}} = \sqrt{\frac{14}{3}} = 2.16...$$

$$\frac{|1|+|2|+|3|}{3} = 2$$

Ok, so we're trying to figure out how much AC voltage we need to dissipate the same amount of average power through a resistive load as DC voltage. I'll use the following equation to calculate the power:

$$P=\frac{V^2}{R}$$

The power through a resistive load with DC: $$P_{DC} = \frac{V_{DC}^2}{R}$$

The instantaneous power through a resistive load with AC: $$P_{AC} = \frac{V_{P}^2}{R}.sin^2\theta$$

The average power through a resistive load with AC: $$\overline{P_{AC}} = \frac{1}{2\pi}\int_{0}^{2\pi} \frac{V_{P}^2}{R}sin^2\theta \,d\theta$$

$$= \frac{V_{P}^2}{2 \pi R}\int_{0}^{2\pi}sin^2\theta \,d\theta$$

$$= \frac{V_{P}^2}{2 \pi R}\int_{0}^{2\pi}\frac{1}{2}-\frac{cos2\theta}{2}\theta \,d\theta$$

$$= \frac{V_{P}^2}{2 \pi R}\left[\frac{\theta}{2}-\frac{sin2\theta}{4}\right]_{0}^{2\pi}$$

$$= \frac{V_{P}^2}{2 \pi R}\pi$$

$$= \frac{V_{P}^2}{2R}$$

The whole point of this RMS stuff is to find how much AC voltage we need to dissipate the same amount of power through a resistive load as DC voltage:

$$\overline{P_{AC}} = P_{DC}$$

$$\frac{V_{P}^2}{2R} = \frac{V_{DC}^2}{R}$$

$$V_{p}^2 = 2 V_{DC}^2$$

$$V_{p} = \sqrt{2}V_{DC}$$

We've arrived at the same conversion formula without any RMS stuff. So why bother with the RMS incantation? If you review the math you'll find that it follows the same pattern as the math for RMS. The square comes from the square in the power equation. I'm not a mathematician (or well versed in the history of maths), but I'm guessing this similarity is where the RMS stuff comes from (RMS is used in many fields). There is no need for any knowledge of RMS to derive the results.

• Good answer and welcome to the site. It's worth noting that you are only considering the case of a sinusoidal waveform and with full cycles. For non-sinusoids such as rectifier currents or SCR controlled AC waveforms the simple $V_p = \sqrt 2 V_{DC}$ equation doesn't hold. Commented Mar 25, 2023 at 17:51

I see all sorts of "proofs" for RMS calculations. RMS is not something that is inherently "correct", it is a useful measure for particular situations.

So let's get down to the details that make it useful for a particular situation: energy generally is interesting as a conserved measure and can often be represented as a time integral of power, and power as a product of two measures.

Electric power at a given moment can be defined as the product of current and voltage. There are lots of situations where current and voltage (or other power-defining measures) are in a proportional relation. When this is the case (and it often is, but one cannot depend on it similarly as with conservation of energy) and we are talking about stationary conditions (like AC circuits that repeat the same energy flow with each cycle), RMS values are convenient for simplifying calculations a whole lot.

This is primarily the case for resistive work loads, but reactive loads (capacities and inductivities) can also benefit from calculations based on RMS values, though in a less direct manner.

That doesn't give RMS values some physical meaning. They are a mathematical tool that is particularly useful in the context of AC calculations.

y(x) = |x| is not differentiable, because y'(0) is undefined.

y(x) = sqrt(x * x) is differentiable.

However they are otherwise equivalent.

Vrms = average(abs(v(t))) = average(sqrt(v(t) * v(t)))

Why did they pick one definition over the other? Well, one is an average of a differentiatable function.

• That's not why, though. It's because using the RMS voltage gives you the same average power as if you calculated the instantaneous power at each point and then averaged it. This also holds for current. All of the equations for DC behavior hold exactly for AC, if and only if the RMS value is used. Commented May 27, 2019 at 0:40
• Not mathamatically correct$$\sqrt{\frac{1^2 + 2^2 + 3^2}{3}} = \sqrt{\frac{14}{3}} = 2.16...$$ $$\frac{|1|+|2|+|3|}{3} = 2$$ Commented Mar 25, 2023 at 16:03